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Vehicle Dynamics and Control
Associate Professor
Dept. Electrical EngineeringVehicular Systems
Linkoping UniversitySweden
Lecture 1
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 1 / 43
Course Literature
Course book is Theory of Ground Vehicles, 4th edition, by J.Y. WongYou can borrow a copy during the course.
Chapter 1: Mechanics of Pneumatic Tires
Chapter 3: Performance Characteristics of Road Vehicles
Chapter 5: Handling Characteristics of Road Vehicles
Chapter 7: Vehicle Ride Characteristics
Some additional material is taken from the books Vehicle Dynamics,Stability and Control, 2nd edition, D. Karnopp, and Tire and VehicleDynamics, H. Pacejka.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 2 / 43
Chapter 1: Mechanics of Pneumatic Tires
ω
V
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 3 / 43
Chapter 3: Performance Characteristics of Road Vehicles
F = ma
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 4 / 43
Chapter 3: Performance Characteristics of Road Vehicles
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 5 / 43
Chapter 5: Handling Characteristics of Road Vehicles
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 6 / 43
Chapter 7: Vehicle Ride Characteristics
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 7 / 43
L-building
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 8 / 43
L-building
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 9 / 43
L-building
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 10 / 43
L-building
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 11 / 43
L-building
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 12 / 43
Stability
Direction of motion
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 13 / 43
Stability
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 14 / 43
Stability
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 15 / 43
Stability
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 16 / 43
Tapered wheels
Why is the wheels on a train tapered?
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 17 / 43
Tapered wheels: Basic motion
Consider a wheelset with tapered wheels on a rail. In the steadymotion/basic motion, the wheels are moving on a straight line in thelongitudinal direction:
2r0
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 18 / 43
Tapered wheels: A train taking a turn
One reason for using tapered wheels is illustrated in the following figureshowing a wheelset of a train taking a right turn:
y
ψ
2r0 + 2ψy 2r0 − 2ψy
The longitudinal speed is larger for the outside wheel Vxl than for theinside wheel Vxr , but the rotational speed ω is the same. The basic motionin this case includes a constant drift y in the lateral direction, whichcompensates for this difference:
Vxl = (r0 + ψy)ω, Vxr = (r0 − ψy)ω
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 19 / 43
Tapered wheels: Perturbed motion
What will happen if the basic motion is perturbed?Basic motion is shown to the left and perturbed motion to the right:
θ ≈ dydx
x
y
ψ
(x(t), y(t))
2r0
Vxr
Vxl
w
ω
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 20 / 43
Tapered wheels
w(x(t), y(t))
Vxr
Vxl
θ ≈ dydx
Lateral drift causes a difference in the longitudinalvelocity of the wheels in the same way as before:
Vxl = (r0 + ψy)ω
Vxr = (r0 − ψy)ω
The longitudinal velocity of the center of gravity isnow given by:
x =Vxl + Vxr
2= r0ω
The approximation dydx ≈ θ gives the lateral velocity:
y =dy
dx
dx
dt= θr0ω
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 21 / 43
Tapered wheels
w(x(t), y(t))
Vxr
Vxl
θ ≈ dydx Using Vxl = (r0 + ψy)ω and Vxr = (r0 − ψy)ω the
angular velocity can be written as
θ =Vxr − Vxl
w= −2ψyω
w
Differentiating y = θr0ω and using the expressionfor the angular velocity above, the followingdifferential equation for y is obtained:
y +2r0ψω
2
wy = 0
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 22 / 43
Tapered wheels: Harmonic oscillation
For a wheelset with positive taper angle (as in the figure) the solution
y(t) +2r0ψω
2
wy(t) = 0
is a harmonic oscillation
y(t) = cos(ωnt + φ)
with natural frequency
ωn =
√2r0ψ
wω
If there is friction in the system, then the wheelset will return to the basicmotion asymptotically.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 23 / 43
Tapered wheels: Unstable system
For a wheelset with negative taper angle the solutions of the differentialequation
y(t) +2r0ψω
2
wy(t) = 0
are
y(t) = C exp
(±√
2r0ψ
wω
)which means that the a small perturbation would cause an exponentialgrowth of the lateral displacement and the system is clearly unstable.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 24 / 43
Tapered wheels: Spatial coordinates
w(x(t), y(t))
Vxr
Vxl
θ ≈ dydx
The dynamic equation
y(t) +2r0ψω
2
wy(t) = 0
can be rewritten by using the relations
y =d2y
dx2x2, ω2 =
x2
r20
and the result is the following:
y ′′(x) +2ψ
wr0y(x) = 0
A model that doesn’t depend on speed.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 25 / 43
Tire
Figure 1.1: Tire construction
Figure 1.2: Coordinates, forces, and moments.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 26 / 43
Rolling resistance
The rolling resistance of tires is primarily caused by the hysteresis in tirematerials due to the deflection of the carcass while rolling.Other less important contributors to the rolling resistance are:
Friction between the tire and the road caused by sliding
Air circulating inside the tire
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 27 / 43
Hysteresis
Exampel of a hysteresis loop caused by friction:
d
Direction of motion
Direction of motion
Displacement
−Ffriction
Ffriction
Ffriction
Ffriction
F
F
F
The energy loss due to hysteresis is equal to the shaded in the figure:
2 · d · Ffriction
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 28 / 43
Rolling resistance: Hysteresis
The center of normal pressure is shifted in the direction of motion due tothe hysteresis
Normal pressure
Deformation
O
Fz
Fr
The applied wheel torque on free-rolling tire is zero. Therefore, ahorizontal force Rr at the contact patch must exists to maintainequilibrium. This force is called known as the rolling resistance.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 29 / 43
Rolling resistance
The coefficient of rolling resistance fr is defined as the ratio of the rollingresistance Rr to the normal load W , i.e., fr = Rr/W .Empirical formulas for calculating the rolling resistance coefficient as afunction of speed V , based on experimental data:
Radial-ply passenger car tire: fr = 0.0136 + 0.40 × 10−7V 2
Radial-ply truck tire: fr = 0.006 + 0.23 × 10−6V 2
Other factors that affect the rolling resistance:
Surface texture, Figure 1.5.
Inflation pressure, Figure 1.7 and 1.8.
Internal temperature, Figur 1.11 and 1.12.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 30 / 43
A Tire Under the Action of a Driving Torque
ω
V
Definitions:
Rolling radius of a free-rolling tire: r = V /ω,
Effective rolling radius under the action of a driving torque: re = V /ω,
where V is the linear speed of the tire center, and ω is the angular speed.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 31 / 43
A Tire Under the Action of a Driving Torque
Longitudinal slip
i =
(1 − V
ωr
)× 100% =
(1 − re
r
)× 100%
Limit cases:
Free-rolling tire: i = 0
The tire is not moving: i = 100% om V = 0,
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 32 / 43
Driving Wheel: The Brush Model
The brush model is a very simple physical model of tire. The tread of thetire is modeled as elastic bristles attached to the rim, and longitudinalforce is generated by the deflection of the brush elements.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 33 / 43
Driving Wheel: The Brush Model
The contact patch is assumed to rectangular and can be divided into anadhesion region (0 ≤ x ≤ lc) and a sliding region (lc ≤ x ≤ lt).
lt
lc
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 34 / 43
Driving Wheel: The Brush Model
The objective is to find the length of the adhesion region lc . When doesthe longitudinal force becomes so large that the bristles begins to slide?
Consider a bristle in the adhesion region
e = ε x
ωr − Vx
The velocity at the rim is ωr − V . The time since the bristle first touchthe ground is t = x/(ωr). The deflection at the distance x is:
e(x) = (ωr − V )x
ωr=
(1 − V
ωr
)x = ix
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 35 / 43
Driving Wheel: The Brush Model
User a linear model for the relation between deflection and longitudinalforce per unit of length:
dFxdx
= kte = kt ix
It is assumed that normal force W is uniformly distributed in the contactregion,
dFzdx
=W
lt
where lt is the length of the contact region.
Assumption: The bristle will not slide if
dFxdx
< µpdFzdx
where µp is the coefficient of friction.
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 36 / 43
Driving Wheel: The Brush Model
The condition can be written
kt ix < µpW
lt
First case: When is there no sliding region?Answer: When x = lt fulfills the condition above, i.e.
kt lt i <µpW
ltor i <
µpW
kt l2t≡ ic
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 37 / 43
Driving Wheel: The Brush Model
The distribution of the longitudinal force in this case (i < ic)
kt lt i
lt
µpWlt
dFxdx
x
Fx = Area of the shaded region =1
2kt l
2t i ≡ Ci i
In the limit case i = ic =µpW
kt l2tis
Fx =1
2kt l
2t
µpW
kt l2t=µpW
2≡ Fxc
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 38 / 43
Driving Wheel: The Brush Model
The second case: There is a sliding region (i > ic).
The distribution of the longitudinal force in this case:
ltlc
µpWlt
dFxdx
x
How do we calculate the length of the adhesion region lc?
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 39 / 43
Driving Wheel: The Brush Model
Solution: Recall that the bristle will not slide if kt ix < µpW /lt , i.e.,
x ≤ µpW
kt lt i≡ lc
ltlc
µpWlt
dFxdx
x
The longitudinal force is equal to the shaded area
Fx =1
2
µpW
ltlc +
µpW
lt(lt − lc) = µpW
(1 − 1
2
lclt
)Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 40 / 43
The Brush Model: Summary
Critical values if longitudinal slip and force:
ic =µpW
kt l2t=µpW
2Cioch Fxc =
µpW
2= Ci ic
There is no sliding region when i ≤ ic eller Fx ≤ Fxc and in this case
Fx =kt l
2t
2i = Ci i
If i > ic eller Fx > Fxc , then the length of the adhesion region is
lc =µpW
kt lt i
and the longitudinal force is
Fx = µpW
(1 − 1
2
lclt
)= µpW
(1 − µpW
4Ci i
)Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 41 / 43
Braking Wheel: The Brush Model
The skid is defined
is =
(1 − ωr
V
)× 100% =
(1 − r
re
)× 100%
when a braking torque is applied to the wheel.
Limit cases:
Free-rolling tire: is = 0
Locked wheel is = 100%
Relations between i and is :
i = − is1 − is
and
is = − i
1 − i
Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 42 / 43
Braking Wheel: Summary
Cs =∂Fx∂is
∣∣∣∣is=0
Critical values of skid and longitudinal force
isc =µpW
2Cs + µpW
Fxc =Cs isc
1 − isc=µpW
2
No slide region (is < isc):
Fx =Cs is
1 − is
With slide region (is ≥ isc):
Fx = µpW
(1 − µpW (1 − is)
4Cs is
)Jan Aslund (Linkoping University) Vehicle Dynamics and Control Lecture 1 43 / 43